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A+B-C=3pi then find sinA+sinB-sinC

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the answer:
the full question is as follow:

if A+B-C=3pi, then find sinA+sinB-sinC

first, the main formula of sine and cosine are:
sinC = 2sin(C/2)cos(C/2)
sinA +sinB = 2sin[(A+B)/2]cos[(A-B)/2]

therefore:
sinA+sinB-sinC = 2sin[(A+B)/2]cos[(A-B)/2] - 2sin(C/2)cos(C/2)

sin[(A+B)/2] = cos(C/2)
2sin(C/2)cos(C/2) = cos[(A+B)/2
and
A+B-C=3 pi implies A+B =3 pi + C, so
cos[(A+B)/2] = cos [3 pi/2 + C/2]
and with the equivalence cos (3Pi/2 + X) = sinX

sinA+sinB-sinC = cos(C/2)+ sin(C/2)
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